INF 556: Topological Data Analysis Fall 2017 Lecture 3: Solution to PC 3-4 Lecturer: Steve Oudot T.A.: Théo Lacombe Disclaimer : Some typos and errors may remain. Please report them to [email protected]. Homology groups of some common spaces All the following computations are done with the eld of coecient Z=2Z, which (among other things) means that we do not care about orientation. The notation E F means that E and F are isomorphic as vector spaces. ' 1. The circle. 2 K = 1 , 2 , 3 0 { } { } { } K = 1, 2 , 1, 3 , 2, 3 1 { } { } { } 1 3 Figure 3.1: Triangulation of the circle We detail in this simple example two approaches to compute homology. Let's begin with the "hand- craft" one: To compute H0, we need Z0 = ker(@0) and B0 = im(@1). One has ker(@0) = span 1 ; 2 ; 3 , ff g f g f gg which are three linearly independent points in our complex (you cannot express 3 as a linear f g combination of 1 and 2 and so on). Therefore: f g f g 3 Z ker(@0) ' 2 Z On the other hand, im(@1) = span 2 1 ; 3 2 ; 1 3 . (Reminder: since we are ff g − f g f g − f g f g − f gg working with Z=2Z, +1 = 1 and thus signs do not matter). However, one can observe that − 1 3 = 2 1 ( 3 2 ), so we actually have im(@1) = span 2 1 ; 3 2 (both vectorsf g − f areg independent)f g − f g − f andg − thus: f g ff g − f g f g − f gg 2 Z im(@1) ' 2 Z and nally: 1 Z Z H0 ; (3.1) S 2 ' 2 Z Z In particular, β0 = 1, which can be interpreted as the circle has one connected component. 3-1 Lecture 3: Solution to PC 3-4 3-2 In order to compute H1, we need to nd the 1-cycles of our complex. One can easily observe that: @1( 1; 2 + 2; 3 + 3; 1 ) = 2 1 + 3 2 + 1 3 f g f g f g f g − f g f g − f g f g − f g = 0 and we do note have any other 1-cycle, so: Z ker(@1) ' 2 Z Aside, we do not have any 2-simplex and so im(@2) = 0 , which nally implies that: f g 1 Z Z H1 ; (3.2) S 2 ' 2 Z Z In particular, β1 = 1, which can be interpreted as there is one non-trivial loop in the circle (which is not a boundary). 1 Z For k 2, Ck ; is trivial and thus so is Hk. > S 2Z Other approach: The idea is the following one: we generally only care about the dimension of Hk (i.e. the Betti number βk), and we have: βk = dim(ker(@k)) rk(@k+1) − We also remind the following fundamental result of linear algebra (in nite dimensional vector spaces): dim(E) = rk(u) + dim(ker(u)) for E a nite dimensional vector space and u a linear application from E to some other vector space. Therefore, we can turn this into the problem of nding the rank of @k (which will also give us the dimension of its kernel), which can be easily computed by writing the matrix of @k: 1, 3 1, 2 { } 2, 3 { } { } 1 1 1 0 {2} 1 0 1 {3} 0 1 1 { } Figure 3.2: Matrix of @ : the starting space is C K; Z , its base is given by the three vectors 1 1 2Z Z 1; 2 ; 2; 3 ; 3; 1 , and we write the coordinates of @1 in the basis 1 ; 2 ; 3 of C0 K; . f g f g f g f g f g f g 2Z Standard computations show that this matrix (whose coecients are in Z=2Z) has rank 2 (see Gaussian elimination): 1 1 0 1 1 1 1 1 0 c3 c3 c2 c c c 1 0 1 ← − 1 0 1 3 ← 3 − 1 1 0 0 0 1 1 0 1 0 0 1 0 Figure 3.3: Sketch of Gaussian elimination to compute the rank of @1 for the circle, leading to a rank 2 matrix. The interest of this algorithm is that it can be easily implemented (and is useful while dealing with more complicated simplices). Lecture 3: Solution to PC 3-4 3-3 2 2. The disk B . 2 K = 1 , 2 , 3 0 { } { } { } K = 1, 2 , 1, 3 , 2, 3 1 { } { } { } K = 1, 2, 3 2 { } 1 3 2 Figure 3.4: Triangulation of B . For H0, computations are exactly the same as the circle (see above). For H1, we have the same result for ker(@1) (one 1-cycle). However, in this case, im(@2) is not empty (we have a 2-simplex), leading to: 2 Z H1 B ; = 0 (3.3) 2 f g Z For H2, despite having a 2-simplex, we do not have any 2-cycle (@2 1; 2; 3 = 1; 2 + 2; 3 + 3; 1 = f g f g f g f g 6 0). Furthermore, we do not have any 3-simplex in this complex, and thus: 2 Z H2 B ; = 0 (3.4) 2 f g Z 2 Finally, B has the same homology groups as a single point, which is actually not a surprise since it is homotopy equivalent to a point! 3. The cylinder c = 1 [0; 1] S × 2 K0 = 1 , 2 , 3 , 4 , 5 , 6 3 { } { } { } { } { } { } 1 K1 = 1, 2 , 1, 3 , 2, 3 , 4, 5 , 4, 6 , 5, 6 ... {... 1}, 5{, 1}, 6{, 2}, 6{, 2}, 4{, 3}, 4{, 3}, 5 6 { } { } { } { } { } { } K = 1, 5, 6 , 1, 6, 2 , 2, 4, 6 , 2, 3, 4 , 1, 5, 3 , 3, 4, 5 2 { } { } { } { } { } { } 2 1 3 2 6 1 3 6 5 4 5 4 5 4 Figure 3.5: Triangulation of the cylinder. The 2-faces 1; 2; 3 and 4; 5; 6 do not belong to the complex. f g f g On the smaller graph, in red, edges such that @1(edges) gives the generators of B0. On the right, a representation of the triangulation "from the top", which can help for computations. Since we have 6 points in this triangulation, dim(Z0) = 6. On the other hand, computations show that im(@1) has 5 (independent) generators (see Fig 3.5). The idea is that the boundary of any other Lecture 3: Solution to PC 3-4 3-4 1-simplex (edge) in the complex can be obtained by going through these edges. For example, 1; 6 f g has 6 1 has a boundary, which can be obtained by taking the boundary of 1; 3 + 3; 4 + 4; 6 . f g−f g f g f g f g Therefore, Z Z H0 c; (3.5) 2 ' 2 Z Z In order to compute H1, we have to nd the 1-cycles and the 1-boundary. There are many 1-cycles in this complex...! For example, any element of the form a; b + b; c + c; a (with a; b ; b; c ; c; a f g f g f g f g f g f g in the complex) is a 1-cycle. However, there are 7 1-cycles (you will nd 8 1-cycles, but one of them can be written as a linear combination of the others), showing that dim(Z1) = 7. On the other hand, we have six 2-faces (triangles). We nally have: Z Z H1 c; (3.6) 2 ' 2 Z Z For H2, we observe that we do not have any 2-cycle, and no 3-simplex, leading to: Z H2 c; = 0 (3.7) 2 f g Z Of course, higher dimensional homology groups are also trivial. Remark: This is the same homology as the circle in question 1. This is not a surprise, since these two spaces are actually homotopy equivalent. 4. The sphere 2 S 4 1 , 2 , 3 , 4 { } { } { } { } 1, 2 , 1, 3 , 2, 3 , 1, 4 , 2, 4 , 3, 4 { } { } { } { } { } { } 1, 2, 3 , 1, 2, 4 , 2, 3, 4 , 1, 3, 4 { } { } { } { } 2 3 1 Figure 3.6: Triangulation of 2. Warning, the 3-face 1; 2; 3; 4 does not belong to the complex. S f g H0 can be computed "by hand" or by computing the rank of the matrix: 1 1 0 1 0 0 1 0 1 0 1 0 @1 = 2 3 0 1 1 0 0 1 60 0 0 1 1 17 6 7 4 5 which are the coordinates of @1(x); x C1 written in the base 1 ; 2 ; 3 ; 4 . This matrix has 2 f g f g f g f g rank 3, and thus β0 = 4 3 = 1, then: − 2 Z Z H0 ; (3.8) S 2 ' 2 Z Z Lecture 3: Solution to PC 3-4 3-5 Similarly, H1 looks at the rank of (boundaries of 2-simplicies written on the base of 1-simplicies): 1 1 0 0 1 0 0 1 21 0 1 03 60 1 0 17 6 7 60 1 1 07 6 7 60 0 1 17 6 7 4 5 which is 3. Since we already know that the rank of @1 is 3, we have dim(ker(@1)) = 6 3 = 3. And − thus, β1 = dim(ker(@1)) rk(@2) = 3 3 = 0. So: − − 2 Z H1 ; = 0 (3.9) S 2 f g Z For H2, we observe that we have one 2-cycle ( 1; 2; 3 + 1; 3; 4 + 1; 2; 4 + 2; 3; 4 ), and no f g f g f g f g 3-simplex.